REPRESENTATION THEORY WEEK 12


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1 REPRESENTATION THEORY WEEK Reflection functors Let Q be a quiver. We say a vertex i Q 0 is +admissible if all arrows containing i have i as a target. If all arrows containing i have i as a source, we call i admissible. By σ i (Q) we denote the quiver obtained from Q by inverting all arrows containing i. Let i be a +admissible vertex and Q = σ i (Q). Let us introduce the functor F + i : Rep Q Rep Q. Let X be a representation of Q. Define X = F + i X as follows. If j i, then X j = X j. Put X i = Kerh, where h = ρ γ : Xj X i, γ=(j i) Q 1 For each γ = (i j) Q define ρ γ : X i X j = X j as the natural projection on the component X j X j. If i is a admissible vertex and Q = σ i (Q) one can define the functor F i : Rep Q Rep Q as follows. Let X = F i (X), where X j = X j for i j, and X i = Coker h, where h = ρ γ : X i X j, γ=(i j) Q 1 and for each γ = (j i) Q define ρ γ: X j = X j X i by restriction of the projection X j Coker h to X j. Example. Let Q be the quiver 1 2, and X is the representation k 0, then F1 (X) = 0 and F 2 + (X) is k k. It is easy to check that F + i is leftexact (maps an injection to an injection) and F i is right exact (maps a surjection to a surjection). Let L i denote the representation of Q which has k in the vertex i and zero in all other vertices. Then F + i (L i ) = 0 and F i (L i ) = 0. Theorem 1.1. Let X be an indecomposable representation of Q and i be a + admissible vertex. Then F + i (X) = 0 iff X = L i. Otherwise X = F + i (X) is indecomposable, (1.1) dimx i = dimx i + dimx j and F i F + i (X) = X. Date: April 27, (j i)
2 2 REPRESENTATION THEORY WEEK 12 If i is admissible vertex and X is indecomposable, then F i (X) = 0 iff X = L i. Otherwise F i (X) is indecomposable, the dimension of F i (X) can be calculated by the same formula (1.1) and F + i F i (X) = X. Proof. NotethatifX = Li, thenhmustbesurjectivebecauseofindecomposabilityof X, hence the formula (1.1) holds. Furthermore, we have the following exact sequence (1.2) 0 X i h (j i) Q 1 X j h X i 0 and ( F i X ) = Coker h = X i i. Observe also that h is injective by definition, hence X is indecomposable. Thus, we proved the first part of the theorem. For the case of admissible vertex, define D: Rep Q Rep Q op, where Q op is the quiver with all arrows of Q reversed, D(X j ) = Xj, D(ρ γ ) = ρ γ, and note that D F + i = F i D. Since D reverses all maps and change Ker to Coker, the second statement of the theorem follows immediately. Note that for an arbitrary X the sequence (1.2) is not exact but h is injective and h h = 0. Therefore one can define a natural injection φ : F i F + i X X, where φ j = id for all j i and φ i coincides with h restricted to Coker h. In the similar way one can define the natural surjection ψ : X F + i F i X if i is a admissible vertex. Finally let Q = σ i (Q), X be a representation of Q and Y be a representation of Q, X = F i X and Y = F + i Y. Let η Hom Q (X,Y ), define χ Hom Q (X,Y) by putting χ j = Id for j i and obtaining χ i from following commutative diagram X i h X h j X i 0 η i η j χ i 0 Y i h Y j h Y i Note χ i is uniquely determined by η. In the same way for each χ Hom Q (X,Y) one can define η Hom Q (X,Y ). A routine check now proves the following Lemma 1.2. Let Q = σ i (Q), X be a representation of Q and Y be a representation of Q, then ( Hom Q F i X,Y ) ( = HomQ X,F + i Y ). 2. Reflection functors and change of orientation. Lemma 2.1. Let Γ be a connected graph without cycles, Q and Q be two quivers on the same graph. Then there exists an enumeration of vertices such that Q = σ k σ 1 (Q) and i is a +admissible vertex for σ i 1 σ 1 (Q).
3 REPRESENTATION THEORY WEEK 12 3 Proof. It is sufficient to prove the statement for two quivers Q and Q different at one arrow. So let γ Q 1. After removing γ, Q splits in two connected components; let Q be the component which contains t(γ). Enumerate vertices of Q in such a way that if i j Q 1, then i > j. This is possible since Q does not have cycles. Check that Q = σ k σ 1 (Q) (here k is the number of all vertices in Q ) and i is a +admissible vertex for σ i 1 σ 1 (Q). Theorem 2.2. Let i be a +admissible vertex for Q and Q = σ i (Q). Then F + i and F i establish a bijection between indecomposable representations of Q (nonisomorphic to L i ) and indecomposable representations of Q (nonisomorphic to L i.) 1 Theorem 2.2 follows from 1.1. Together with Lemma 2.1 it allows to change an orientation on a quiver if the quiver does not have cycles. 3. Weyl group and reflection functors. Given any graph Γ, one can associate with it a certain linear group, which is called aweylgroupofγ. Wedenotebyα 1,...,α n vectorsinthestandardbasisofz Γ 0 = Z n, α i corresponds to the vertex i. These vectors are called simple roots. For each simple root α i put r i (x) = x 2(x,α i) (α i,α i ) α i. One can check that r i preserves the scalar product and r 2 i = id. The linear transformation r i is called a simple reflection. If Γ has no loops, r i also preserves the lattice generated by simple roots. Hence r i maps roots to roots. If Γ is Dynkin, the scalar product is positivedefinite, and r i is a reflection in the hyperplane orthogonal to α i. The Weyl group W is a group generated by r 1,...,r n. For a Dynkin diagram W is finite (since the number of roots is finite). Example. Let Γ = A n. Let ε 1,...,ε n+1 be an orthonormal basis in R n+1. Then onecantaketherootsofγtobeε i ε j, simplerootstobeε 1 ε 2,ε 2 ε 3,...,ε n ε n+1, r i (ε j ) = 0 if j i,i + 1, and r i (ε i ) = ε i+1. Therefore W is isomorphic to the permutation group S n+1. One can check by direct calculation, that (1.1) implies Lemma 3.1. If X is an indecomposable representation of Q and dimx = x α i, then dimf ± i X = r i (x). An element c = r 1...r n W is called a Coxeter transformation. It depends on the enumeration of simple roots. Example. In the case Γ = A n a Coxeter element is always a cycle of length n+1. Lemma 3.2. If c(x) = x, then (x,α i ) = 0 for all i. In particular for a Dynkin graph c(x) = x implies x = 0. 1 We will denote by the same letter L i the representations of quivers with different orientation.
4 4 REPRESENTATION THEORY WEEK 12 Proof. By definition, c(x) = x+a 1 α 1 + +a n α n, a i = 2(α i,x+a 1 α a i 1 α i 1 ). (α i,α i ) The condition c(x) = x implies all a i = 0. Hence (x,α i ) = 0 for all i. 4. Coxeter functor. Let Q be a graph without oriented cycles. We call an enumeration of vertices admissible if i > j for any arrow i j. Such an enumeration always exists. One can easily see that every vertex i is a +admissible for σ i 1 σ 1 (Q) and admissible for σ i+1 σ n (Q). Furthermore, Define Coxeter functors Q = σ n σ n 1 σ 1 (Q) = σ 1 σ n (Q). Φ + = F + n F + 2 F + 1, Φ = F 1 F 2 F n. Lemma 4.1. (1) Hom Q (Φ X,Y) = Hom Q (X,Φ + Y); (2) If X is indecomposable and Φ + X 0, then Φ Φ + X = X; (3) If X is indecomposable of dimension x and Φ + X 0, then dimφ + X = c(x); (4) If Q is Dynkin, then for any indecomposable X there exists k such that (Φ + ) k X = 0. Proof. (1) follows from Lemma 1.2, (2) follows from Theorem 1.1, (3) follows from Lemma 3.1. Let us prove (4). Since W is finite, c has finite order h. It is sufficient to show that for any x there exists k such that c k (x) is not positive. Assume that this is not true. Then y = x+c(x)+ +c h 1 (x) > 0 is c invariant. Contradiction with Lemma 3.2. Lemma 4.2. Φ ± does not depend on a choice of admissible enumeration. Proof. Note that if i and j are disjoint and both +( )admissible, then F + i F + j = ( ) F + j F + i F i F j = F j F i. If a sequence i1,...,i n gives another admissible enumeration of vertices, and i k = 1, then 1 is disjoint with i 1,...,i k 1, hence F + 1 F + i k 1 F + i 1 = F + i k 1 F + i 1 F + 1. Now proceed by induction. Similarly for Φ. In what follows we always assume that an enumeration of vertices is admissible. Corollary 4.3. Let Q be a Dynkin quiver, X be an indecomposable representation of dimension x, and k be the minimal number such that c k+1 (x) is not positive. There exists a unique vertex i such that x = c k r 1...r i 1 (α i ), X = ( Φ ) k F 1 F i 1 (L i).
5 REPRESENTATION THEORY WEEK 12 5 In particular, x is a positive root and for each positive root x, there is a unique (up to an isomorphism) indecomposable representation of dimension x. Proof. Follows from Theorem 1.1 and Lemma Further properties of Coxeter functors Here we assume again that Q is a quiver without oriented cycles and the enumeration of vertices is admissible. We discuss the properties of the bilinear form,. Since we plan to change an orientation of Q we use a subindex, Q, where it is needed to avoid ambiguity. Lemma 5.1. Let i be a +admissible vertex, Q = σ i (Q), and, Q,, Q the corresponding bilinear forms. Then r i (x),y Q = x,r i (y) Q. Proof. It suffices to check the formula for a subquiver containing i and all its neighbors. Let x = r i (x) and y = r i (y). Then x i = x i + i j x j, y i = y i + i j y j, x,y Q = x iy i x i y j + x j y j = x iy i + i j i j i j x,y Q = x i y i y i x j = x iy i + x j y j. i j i j Corollary 5.2. For a Coxeter element c we have c 1 (x),y = x,c(y). x j y j, If Φ + (Y) 0,Φ (X) 0, then dimext 1( X,Φ + (Y) ) = dimext 1( Φ (X),Y ). Proof. First statement follows directly from Lemma 5.1. The second statement follows from the first statement, Lemma 1.2 and the identity x,y Q = dimhom Q (X,Y) dimext 1 (X,Y). Let A = k(q) be the path algebra. Recall that any indecomposable projective module is isomorphic to Ae i. Lemma 5.3. F + i F + 1 (Ae i ) = 0, F + i 1 F+ 1 (Ae i ) = L i.
6 6 REPRESENTATION THEORY WEEK 12 Proof. One can check by direct calculation that for each component e j Ae i, e j Ae i = 0 for j > i, and F + k F+ 1 (e j Ae i ) = e j Ae i for k < j, F + j F + 1 (e j Ae i ) = 0. Corollary 5.4. Φ + (P) = 0 for any projective module P. For any indecomposable projective Ae i we have (5.1) Ae i = F 1 F i 1 (L i). Proof. The first statement follows from Lemma 5.3 immediately. For the second use Theorem 1.1 and Lemma 5.3. An injective module is a module I such that for any injective homomorphism i : X Y and any homomorphism ϕ : X I, there exists a homomorphism ψ : Y I such that ϕ = ψ i. A module I is injective iff Ext 1 (X,I) = 0 for any X. One can see analogy with projective modules, however in general there is no nice description of injective (like a summand of a free module). Exercise. Check that Q is an injective Zmodule. In case when A is a finitedimensional algebra, injective modules are easy to describe. Indeed, the functor D : A mod mod A such that D(X) = X maps left projective modules to right injective and vice versa. Therefore any indecomposable injective module is isomorphic to (e j A). Since D Φ + = Φ D, one can see easily that Φ (I) = 0 for any injective module I. Moreover, one can prove similarly to the projective case that (e j A) = F + n F + j+1 (L j). Let P (j) = Ae j and I(j) = (e j A) and p(j) = dimp (j), i(j) = dimi(j). Then (5.2) c(p(j)) = r n...r 1 (p(j)) = r n...r j+1 ( α j ) = i(j). Note that Ext 1 (Ae j,x) = 0 for any X and dimhom Q (Ae j,x) = x j. Hence (5.3) p(j),x = x j. On the other hand, Ext 1 (X,(e j A) ) = 0 and Hom Q (X,(e j A) ) = Hom Q (e j A,X ), which implies dimhom Q (X,(e j A) ) = x j. Thus, we obtain (5.4) x,i(j) = x j. Combine together (5.2), (5.3), (5.4) and get p(j),x + x,c(p(j)) = 0.
7 REPRESENTATION THEORY WEEK 12 7 Since p(1),...,p(n) form a basis, the last equation implies that for arbitrary x and y (5.5) y,x + x,c(y) = Affine root system Let Γ be an affine Dynkin graph. Then the kernel of bilinear symmetric form in Z n is onedimensional and generated by δ = a 0 α 0 +a 1 δ 1 + +a n δ n. We assume without loss of generality that the vertex α 0 is such that a 0 = 1. By removing 0 from Γ we get a Dynkin graph which we denote by Γ 0. In affine case roots can be of two kinds: real, if q(α) = 1, or imaginary, q(α) = 0. Lemma 6.1. Imaginary roots are all proportional to δ, real roots can be written as α+mδ for some root δ of Γ 0. Every real root can be obtained from a simple root by the action of the Weyl group W. Proof. The first statement is obvious, the second follows from the fact that q(α) = q(α+mδ), hence the projection on the hyperplane generated by α 1,...α n maps a root to a root. To prove the last statement, note that r i maps every positive root differentfromα i toapositiveroot. Letαbeapositiverealroot, α = a 0 α 0 + +a n α n, and h(α) = a 0 + a a n. Then (α,α i ) > 0 at least for one i. But then h(r i (α)) < h(α). Thus, one can decrease h(α) by application of simple reflection. Intheendonecangetarootofheight1, whichisasimpleroot. Similarlyfornegative roots. 7. Kronecker quiver In this section we use Coxeter functors to classify indecomposable representation of the quiver Â1 =. The admissible enumeration of vertices is 1 0, δ = α 0 +α 1. Positive real roots are mα 1 +(m+1)α 0 = α 1 +(m+1)δ, (m+1)α 1 +mα 0 = α 1 +mδ, m 0. The Coxeter element c = r 1 r 0 satisfies c(α 1 ) = α 1 +2δ, c(δ) = δ. Letx = mα 1 +lδ. Ifm > 0thenc s (x)isnotpositiveforsufficientlylarges. Henceif X is indecomposable of dimension x, then (Φ ) s X = 0. If m < 0, then (Φ + ) s X = 0. Thus if m 0, then as in the case of Dynkin quiver, X can be obtained from some L i by application of reflection functor. In particular, we obtain that the dimension of an indecomposable representation is always a root and if this root is real, then the indecomposable with this dimension is unique up to an isomorphism. Indeed, we have either k m A B k m+1,
8 8 REPRESENTATION THEORY WEEK 12 where A = (1 m,0),b = (0,1 m ), or k m+1 C D k m, where C = A t,d = B t. Classification of indecomposables of dimension mδ is equivalent to classification of pairsoflinearoperators(a,b) : k m k m uptoequivalence(a,b) (PAQ 1,PBQ 1 ). AssumethatAisinvertible, thenonemayassumethata = Id, andthenclassifyb up to conjugation. Indecomposability of the representation implies that B is equivalent to the Jordan block with some eigenvalue µ. Denote the corresponding representation by ρ µ. If B is invertible, then A is equivalent to a Jordan block. Denote such representation by σ µ. One can see that ρ µ is isomorphic to σ µ 1 if µ 0. Now let us prove that at least one of A and B is invertible. Indeed, indecomposability implies that KerA KerB = 0. Hence A+tB is invertible for some t. Without loss of generality one can assume that A+tB = id. But then either A or B must be invertible. Thus, we proved that indecomposable representation of dimension (m, m) = δ are parameterized by a projective line. For other affine quivers, the situation is more complicated, as there are real roots which remain positive under Coxeter transformation. For example consider the quiver D Then c(α 1 +α 2 +α 3 ) = α 4 +α 1 +α 5, c 2 (α 1 +α 2 +α 3 ) = α 1 +α 2 +α 3.
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